Problem 39
Question
Solve each equation by first finding the LCD for the fractions in the equation and then multiplying both sides of the equation by it.(Assume \(x\) is not 0 in Problems \(39-46\).) $$\frac{4}{x}=\frac{1}{5}$$
Step-by-Step Solution
Verified Answer
The solution is \(x = 20\).
1Step 1: Identify the Least Common Denominator (LCD)
To solve the equation \(\frac{4}{x} = \frac{1}{5}\), first identify the Least Common Denominator (LCD) of the fractions. The denominators are \(x\) and \(5\). Since \(x\) is variable, the LCD is simply \(5x\).
2Step 2: Multiply Both Sides by the LCD
Multiply both sides of the equation by the LCD, \(5x\), to eliminate the fractions: \[ 5x \cdot \frac{4}{x} = 5x \cdot \frac{1}{5} \] This simplifies to: \[ 20 = x \]
3Step 3: Verify the Solution
To verify, substitute \(x = 20\) back into the original equation: \[ \frac{4}{20} = \frac{1}{5} \] Simplify the left side: \[ \frac{1}{5} = \frac{1}{5} \] Both sides of the equation are equal, confirming that the solution \(x=20\) is correct.
Key Concepts
Understanding the Least Common DenominatorWorking with FractionsVerifying Solutions
Understanding the Least Common Denominator
When it comes to equations involving fractions, one essential step is finding the "Least Common Denominator" (LCD). The LCD is the smallest number that both denominators can divide into without leaving a remainder. This technique is valuable because it allows you to clear the fractions and work with whole numbers, which are easier to manage.
For example, in the equation \( \frac{4}{x} = \frac{1}{5} \), the denominators are \( x \) and 5. Here, \( x \) is a variable, so the LCD is just the product of these two terms: \( 5x \). By using the LCD, you essentially level the playing field, making it simpler to solve the equation.
Finding the LCD is particularly beneficial:
For example, in the equation \( \frac{4}{x} = \frac{1}{5} \), the denominators are \( x \) and 5. Here, \( x \) is a variable, so the LCD is just the product of these two terms: \( 5x \). By using the LCD, you essentially level the playing field, making it simpler to solve the equation.
Finding the LCD is particularly beneficial:
- It simplifies the equation by getting rid of fractions.
- It provides a consistent baseline for adjusting each term of the equation.
Working with Fractions
Fractions can often seem daunting to work with, especially in equations. However, understanding their composition helps in simplifying these equations.
A fraction like \( \frac{4}{x} \) consists of two parts:
When fractions are involved, remember to:
A fraction like \( \frac{4}{x} \) consists of two parts:
- The numerator (above the line), which is 4 in this case.
- The denominator (below the line), which is the variable \( x \).
When fractions are involved, remember to:
- Identify the numerators and denominators first.
- Use the LCD to transform the equation into one without fractions.
- Be cautious with any variable in the denominator, as they imply restrictions on the values the variable can take (like \( x eq 0 \) if \( x \) is in the denominator).
Verifying Solutions
Once you have solved an equation, the final crucial step is to verify its solution. Verifying essentially checks whether the value you found actually satisfies the original equation.
For example, if you obtain \( x = 20 \) from \( \frac{4}{x} = \frac{1}{5} \), you substitute \( 20 \) back into the original equation to ensure it holds true:
For example, if you obtain \( x = 20 \) from \( \frac{4}{x} = \frac{1}{5} \), you substitute \( 20 \) back into the original equation to ensure it holds true:
- Substituting back gives \( \frac{4}{20} \).
- When simplified, \( \frac{4}{20} = \frac{1}{5} \), which matches the right side of the equation.
- It confirms the correctness of your calculations and reasoning.
- It reveals any potential mistakes made during the process, allowing for corrections.
Other exercises in this chapter
Problem 39
Using the addition property of equality first, solve each of the following equations. $$-2 x-5=-7$$
View solution Problem 39
Apply the distributive property to each expression and then simplify. $$6(4 y-3)+6 y$$
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On a certain day, the temperature on the ground is \(72^{\circ}\) Fahrenheit, and the temperature \(T\) at an altitude of \(A\) feet above the ground is given b
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The following equations contain parentheses. Apply the distributive property to remove the parentheses, then simplify each side before using the addition proper
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